4u² + 8u = 0
4u(u + 2) = 0
u = 0 or u = -2

Step-by-step explanation:

Taking q as some integer.

Let a be the positive integer.

And, b = 4 .

Then by Euclid's division lemma,

We can write a = 4q + r ,for some integer q and 0 ≤ r < 4 .

°•° Then, possible values of r is 0, 1, 2, 3 .

Taking r = 0 .

a = 4q .

Taking r = 1 .

a = 4q + 1 .

Taking r = 2

a = 4q + 2 .

Taking r = 3 .

a = 4q + 3 .

But a is an odd positive integer, so a can't be 4q , or 4q + 2 [ As these are even ] .

•°• a can be of the form 4q + 1 or 4q + 3 for some integer q .

Hence , it is solved

Let us suppose that 'a' be the first term and 'd' be the common difference of the A.P.
Now according to question it is given that

S₉ = 162.

Using formula for sum to n terms of A.P, we have
{tex}\Rightarrow \frac{9}{2}\left[ {2a + (9 - 1)d} \right] = 162{/tex}
{tex} \Rightarrow \frac{9}{2}\left[ {2a + 8d} \right] = 162{/tex}
{tex}\Rightarrow{/tex} 9a + 36d = 162 ........................(i)
Let a₆ and a₁₃ be the 6^_th and 13^th term of the A.P. respectively.
Therefore, using general form of nth term, we have,a₆ = a + 5d and a₁₃ = a + 12d
Since, {tex}\frac{{{a_6}}}{{{a_{13}}}} = \frac{1}{2}{/tex}
{tex} \Rightarrow \frac{{a + 5d}}{{a + 12d}} = \frac{1}{2}{/tex}
{tex}\Rightarrow{/tex} 2a + 10d = a + 12d
{tex}\Rightarrow{/tex} a = 2d
Now,substituting the value of a = 2d in equation (i), we get,
9(2d) + 36d = 162
{tex}\Rightarrow{/tex} 18d + 36d = 162
{tex}\Rightarrow{/tex} 54d = 162
{tex}\Rightarrow{/tex} d = 3
{tex}\Rightarrow{/tex} a = 2d = 2 {tex}\times{/tex} 3 = 6 = First term

Therefore, 15th term=a+14d=6+14(3)=48

n³ - n = n (n² - 1) = n (n - 1) (n + 1) 

Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.
If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.
⇒ n (n – 1) (n + 1) is divisible by 3.
 
Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.
∴ n = 2q or 2q + 1, where q is some integer.
If n = 2q, then n is divisible by 2.
If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.
⇒ n (n – 1) (n + 1) is divisible by 2.
Since, n (n – 1) (n + 1) is divisible by 2 and 3.

∴ n (n-1) (n+1) = n³ - n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6) 
 

Let the speed of the boat in still water be x km/hr.
Speed of the stream = 2 km/hr.
{tex}\therefore{/tex} speed downstream = (x + 2) km/hr,
speed upstream = (x - 2) km/hr.
Time taken by boat to go 8 km downstream = {tex}\frac{8}{x+2} {\ hr}{/tex} 

Time taken by boat to return 8 km upstream = {tex}\frac{8}{x-2} hr{/tex} 

Total time taken by boat = 1 hr and 40 minutes

According to question,
{tex}\therefore \quad \frac { 8 } { x + 2 } + \frac { 8 } { x - 2 } = \frac { 5 } { 3 }{/tex}
{tex}\Rightarrow \frac { 1 } { x + 2 } + \frac { 1 } { x - 2 } = \frac { 5 } { 24 } \Rightarrow \frac { ( x - 2 ) + ( x + 2 ) } { ( x + 2 ) ( x - 2 ) } = \frac { 5 } { 24 }{/tex}
{tex}\Rightarrow \frac { 2 x } { \left( x ^ { 2 } - 4 \right) } = \frac { 5 } { 24 } \Rightarrow{/tex} 5x² - 20 = 48x
{tex}\Rightarrow{/tex} 5x² - 48x - 20 = 0

{tex}\Rightarrow{/tex} 5x² - 50x + 2x - 20 = 0
{tex}\Rightarrow{/tex} 5x(x - 10) + 2(x - 10) = 0

{tex}\Rightarrow{/tex}(x - 10)(5x + 2) = 0
{tex}\Rightarrow{/tex} x - 10 = 0 or 5x + 2 = 0
{tex}\Rightarrow{/tex} x = 10 or x = {tex}\frac { - 2 } { 5 }{/tex}
{tex}\Rightarrow{/tex} x = 10 [{tex}\because{/tex} speed can't be negative]
Therefore, the speed of the boat is 10 km/hr.

Let the starting salary of the man be Rs. x and the fixed annual increment be Rs.y.
Then,
Salary after 4 years of service = Rs. (x + 4y)
Salary after 10 years of service = Rs. (x + 10y)
As per given condition
After four years salary was Rs.1500
  x + 4y =1500 .............. (i)
And Rs.1800 after 10 years of service.
So, x + 10y = 1800 .............. (ii)
Subtracting equation (i) from equation (ii), we get
(x + 10y) -  (x + 4y) = 1800 - 1500
{tex}\Rightarrow{/tex} x + 10y - x - 4y = 300
{tex}\Rightarrow{/tex} 6y = 300
 {tex}\Rightarrow{/tex} y =50
Putting y = 50 in equation (i), we get
x + 4(50) = 1500
x = 1300.
Hence the starting salary was Rs. 1300 and annual increment is Rs.50.

You can check NCERT Solutions here : https://mycbseguide.com/ncert-solutions.html

{tex}A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&1 \\ 2&1&{ - 3} \\ 1&1&1 \end{array}} \right]{/tex}

{tex}\left| A \right| = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&1 \\ 2&1&{ - 3} \\ 1&1&1 \end{array}} \right]{/tex}

{tex} = 10 \ne 0{/tex}

Here,
A₁₁ = 4, A₁₂ = -5, A₁₃ = 1
A₂₁ = 2, A₂₂ = 0, A₂₃ = -2
A₃₁ = 2, A₃₂ = 5, A₃₃ = 3

{tex}adjA = \left[ {\begin{array}{*{20}{c}} 4&2&2 \\ { - 5}&0&5 \\ 1&{ - 2}&3 \end{array}} \right]{/tex}

{tex}{A^{ - 1}} = \frac{1}{{\left| A \right|}}\left( {adjA} \right){/tex}

{tex}= \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} 4&2&2 \\ { - 5}&0&5 \\ 1&{ - 2}&3 \end{array}} \right]{/tex}

System of equation can be written is

X = A^-1B

{tex} = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} 4&2&2 \\ { - 5}&0&5 \\ 1&{ - 2}&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 4 \\ 0 \\ 2 \end{array}} \right]{/tex}

{tex}\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right]{/tex}

x = 2, y = -1, z = 1

Let a = 4q + r, when r = 0, 1, 2 and 3
{tex}\therefore{/tex}Numbers are 4q, 4q + 1, 4q + 2 and 4q + 3
{tex}( a ) ^ { 2 } = ( 4 q ) ^ { 2 } = 16 q ^ { 2 } = 4 ( 4 q ) ^ { 2 } = 4 m{/tex}
{tex}( a ) ^ { 2 } = ( 4 q + 1 ) ^ { 2 } = 16 q ^ { 2 } + 8 q + 1 = 4 \left( 4 q ^ { 2 } + 2 q \right) + 1 = 4 m + 1{/tex}
{tex}( a ) ^ { 2 } = ( 4 q + 2 ) ^ { 2 } = 16 q ^ { 2 } + 16 q + 4 = 4 \left( 4 q ^ { 2 } + 4 q + 1 \right) = 4 m{/tex}
{tex}( a ) ^ { 2 } = ( 4 q + 3 ) ^ { 2 } = 16 q ^ { 2 } + 24 q + 9 = 4 \left( 4 q ^ { 2 } + 6 q + 2 \right) + 1 = 4 m + 1{/tex}
{tex}\therefore {/tex} the square of any +ve integer is of the form 4q or 4q + 1 

We know that, cosec30°=2, cos60°=(1/2), tan45°=1=sin90°, sec45°=√2 & cot30°=√3,
putting these values in the given expression, we get:-
{tex}\cos e{c^3}30^\circ \cos 60^\circ {\tan ^3}45^\circ {\sin ^2}90^\circ {\sec ^2}45^\circ \cot 30^\circ {/tex}
{tex} = {(2)^3} \times {\left( {\frac{1}{2}} \right)} \times {(1)^3} \times {(1)^2} \times {\left( {\sqrt 2 } \right)^2} \times \sqrt 3 {/tex}
{tex} = 8 \times \frac{1}{2} \times 1 \times 1 \times 2 \times \sqrt 3 {/tex}
{tex} = 8\sqrt 3 {/tex}